JEE Advanced 2022 Paper 1 · Q14 · Determinants

Question

Let $|M|$ denote the determinant of a square matrix $M$. Let $g:[0,\pi/2]	o\mathbb{R}$ be the function defined by $g(	heta)=\sqrt{f(	heta)-1}+\sqrt{f(\pi/2-	heta)-1}$ where $f(	heta)=\dfrac{1}{2}\begin{vmatrix}1 & \sin	heta & 1\-\sin	heta & 1 & \sin	heta\-1 & -\sin	heta & 1\end{vmatrix}+\begin{vmatrix}\sin\pi & \cos(	heta+\pi/4) & 	an(	heta-\pi/4)\\sin(	heta-\pi/4) & -\cos(\pi/2) & \log_{e}(4/\pi)\\cot(	heta+\pi/4) & \log_{e}(\pi/4) & 	an\pi\end{vmatrix}$. Let $p(x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(	heta)$, and $p(2)=2-\sqrt{2}$. Then, which of the following is/are TRUE?

SolveKar AI · Accepted Answer

First determinant: expanding the $3 imes 3$ matrix gives $1\cdot(1+\sin^{2} heta)-\sin heta(-\sin heta+\sin heta)+1\cdot(\sin^{2} heta+1)=2(1+\sin^{2} heta)$, so its half $=1+\sin^{2} heta$. Second determinant: first row is $(0,\cos( heta+\pi/4), an( heta-\pi/4))$; second row has $-\cos(\pi/2)=0$ and $\log_{e}(4/\pi)$; third row has $ an\pi=0$. Computing yields a constant. Combined, $f( heta)=1+\sin^{2} heta+ ext{const}$. After algebra (skipped here for space), $g( heta)=|\sin heta|+|\cos heta|$ on $[0,\pi/2]$. Max $=\sqrt{2}$ at $ heta=\pi/4$, min $=1$ at endpoints. So $p(x)=k(x-1)(x-\sqrt{2})$. $p(2)=k(1)(2-\sqrt{2})=2-\sqrt{2}\Rightarrow k=1$. So $p(x)=(x-1)(x-\sqrt{2})$; negative on $(1,\sqrt{2})$, positive outside. Check: $(3+\sqrt{2})/4\approx 1.10\in(1,\sqrt{2})\Rightarrow p<0$ (A TRUE). $(1+3\sqrt{2})/4\approx 1.31\in(1,\sqrt{2})\Rightarrow p<0$ (B false). $(5\sqrt{2}-1)/4\approx 1.52>\sqrt{2}\Rightarrow p>0$ (C TRUE). $(5-\sqrt{2})/4\approx 0.90<1\Rightarrow p>0$ (D false).

JEE Advanced 2022 Paper 1 · Q14 · Determinants

Solution